LeetCode 214. Shortest Palindrome

题目

Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.

For example:

Given “aacecaaa”, return “aaacecaaa”.

Given “abcd”, return “dcbabcd”.
给你一个串。求最少在前面补多少个字母，能使其成为回文串。

题解

网上基本做法就是构造一个串s+#+反转（s），然后求新串的最长的和后缀相等的前缀。
然后利用kmp里面的next数组正好能求解。
其实这题还有个黑科技。。。
预处理字符串后，能在O(1)的时间里比较两子串是否相同。
链接：HASH求两子串是否相等(O(N)预处理O(1)比较)
所以就变成了很暴力的解法了。。

代码

这里只给出hash的解法，kmp的解法网上到处都是const int MAXN = 50100;const unsigned long long SEED = 13331;unsigned long long P[MAXN];unsigned long long S[MAXN];unsigned long long RS[MAXN];string ts;int n;unsigned long long HS(int l, int r) {//正向    if (l == r) {        return ts[l];    }    l++;    r++;    return S[r] - S[l - 1] * P[r - l + 1];}unsigned long long RHS(int l, int r) {//反向    if (l == r) {        return ts[l];    }    l = n - l - 1;    r = n - r - 1;    swap(l, r);    l++;    r++;    return RS[r] - RS[l - 1] * P[r - l + 1];}string shortestPalindrome(string s) {    ts = s;    P[0] = 1;    for (int i = 1; i < MAXN; i++) {        P[i] = P[i - 1] * SEED;    }    n = s.size();    for (int i = 1; i <= n; i++) {        S[i] = S[i - 1] * SEED + s[i - 1];    }    for (int i = 1; i <= n; i++) {        RS[i] = RS[i - 1] * SEED + s[n - i];    }    int ans = n - 1;    for (int i = 0; i < n / 2; i++) {        if (HS(0, i) == RHS(i + 1, i + 1 + i)) {            ans = min(ans, n - 2 * (i + 1 ));        }        if (2 * i + 2 < n && HS(0, i) == RHS(i + 2, i + 2 + i)) {            ans = min(ans, n - 2 * (i + 1) - 1);        }    }    string ss = "";    for (int i = 0; i < ans; i++) {        ss += s[n - i - 1];    }    ss += s;    return ss;}